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a) ĐKXĐ: \(x\ne3;x\ne\pm2\)
\(C=\frac{2a-a^2}{a+3}\cdot\left(\frac{a-2}{a+2}-\frac{a+2}{a-2}+\frac{4a^2}{4-a^2}\right)\)
\(C=\frac{-a^2+2a}{a+3}\cdot\left(-\frac{4a}{a-2}\right)\)
\(C=-\frac{2a-a^2}{a+3}\cdot\frac{4a}{a-2}\)
\(C=-\frac{\left(2a-a^2\right)\cdot4a}{\left(a+3\right)\left(a-2\right)}\)
\(C=\frac{4a^2}{a+3}\)
b) \(C=\frac{4.4^2}{4+3}=\frac{46}{7}\)
c) \(\frac{4a^2}{a+3}=1\)
<=> 4a2 = a + 3
<=> 4a2 - a - 3 = 0
<=> 4a2 - 3a - 4a - 3 = 0
<=> a(4a + 3) - (4a + 3) = 0
<=> (4a + 3)(a - 1) = 0
<=> 4a + 3 = 0 hoặc a - 1 = 0
<=> a = -3/4 hoặc a = 1
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
để A xác định
\(\Rightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x^2\ne4\end{cases}}\Rightarrow x\ne\pm2\)
\(A=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}\)
\(A=\frac{4.x-8}{\left(x+2\right).\left(x-2\right)}+\frac{3.x+6}{\left(x-2\right).\left(x+2\right)}-\frac{5x-6}{\left(x-2\right).\left(x+2\right)}\)
\(A=\frac{4x-8+3x+6-5x+6}{\left(x+2\right).\left(x-2\right)}=\frac{2.\left(x+2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{2}{x-2}\)
\(\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4x-8}{\left(x+2\right)\left(x-2\right)}+\frac{3x+4}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}=\frac{4x-8+3x+4-5x+6}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{2x+2}{\left(x+2\right)\left(x-2\right)}=\frac{2x+2}{x^2-4}\)
C, \(x=4\Rightarrow A=\frac{2x+2}{x^2-4}=\frac{-6}{12}=\frac{-1}{2}\)
d, \(A\inℤ\Leftrightarrow2x+2⋮x^2-4\Leftrightarrow2x^2+2x-2x^2+8⋮x^2-4\Leftrightarrow2x+8⋮x^2-4\)
\(\Leftrightarrow2x^2+8x⋮x^2-4\Leftrightarrow16⋮x^2-4\)
\(x^2-4\inℕ\)
\(\Rightarrow x^2\in\left\{0;4;12\right\}\)
Thử lại thì 12 ko là số chính phương vậy x=0 hoặc x=2 thỏa mãn
mk học lớp 6 mong mn thông cảm nếu có sai sót
a) \(ĐKXĐ:\hept{\begin{cases}a\ne-3\\a\ne\pm2\end{cases}}\)
\(M=\frac{2a-a^2}{a+3}\left(\frac{a-2}{a+2}-\frac{a+2}{a-2}+\frac{4a^2}{4-a^2}\right)\)
\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{\left(a-2\right)^2-\left(a+2\right)^2-4a^2}{\left(a-2\right)\left(a+2\right)}\)
\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{a^2-4a+4-a^2-4a-4-4a^2}{\left(a-2\right)\left(a+2\right)}\)
\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{-4a^2-8a}{\left(a-2\right)\left(a+2\right)}\)
\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{-4a\left(a+2\right)}{\left(a-2\right)\left(a+2\right)}\)
\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{-4a}{a-2}\)
\(\Leftrightarrow M=\frac{4a^2\left(a-2\right)}{\left(a+3\right)\left(a-2\right)}\)
\(\Leftrightarrow M=\frac{4a^2}{a+3}\)
b) Để M = 1
\(\Leftrightarrow\frac{4a^2}{a+3}=1\)
\(\Leftrightarrow4a^2=a+3\)
\(\Leftrightarrow4a^2-a-3=0\)
\(\Leftrightarrow\left(4a+3\right)\left(a-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4a+3=0\\a-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a=-\frac{3}{4}\left(tm\right)\\a=1\left(tm\right)\end{cases}}\)
Vậy để \(M=1\Leftrightarrow a\in\left\{-\frac{3}{4};1\right\}\)
c) Để M > 0
\(\Leftrightarrow\frac{4a^2}{a+3}>0\)
\(\Leftrightarrow a+3>0\)(Vì 4a2 > 0, loại trường hợp = 0)
\(\Leftrightarrow a>-3\)
Vậy để \(M>0\Leftrightarrow a>-3\)
Để M < 0
\(\Leftrightarrow\frac{4a^2}{a+3}< 0\)
\(\Leftrightarrow a+3< 0\)(Vì 4a2 > 0, loại trường hợp = 0)
\(\Leftrightarrow a< -3\)
Vậy để \(M< 0\Leftrightarrow a< -3\)
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{x^2-9}\)
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
B xác định \(\Leftrightarrow\hept{\begin{cases}x-3\ne0\\x+3\ne0\end{cases}\Leftrightarrow}x\ne\pm3\)
Vậy B xác định \(\Leftrightarrow x\ne\pm3\)
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{x^2-9}\)
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{5\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{5x-15+3x+9-5x-3}{\left(x+3\right)\left(x-3\right)}\)
\(B=\frac{3x-9}{\left(x+3\right)\left(x-3\right)}\)
\(B=\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(B=\frac{3}{x+3}\)